Practice Questions for Recursion | Set 7

Question 1 Predict the output of the following program. What does the following fun() do in general? 

#include <iostream>
using namespace std;

int fun(int n, int* fp)
{
    int t, f;

    if (n <= 2) {
        *fp = 1;
        return 1;
    }
    t = fun(n - 1, fp);
    f = t + *fp;
    *fp = t;
    return f;
}

int main()
{
    int x = 15;
    cout << fun(5, &x) << endl;
    return 0;
}

#include <stdio.h>

int fun(int n, int* fp)
{
    int t, f;

    if (n <= 2) {
        *fp = 1;
        return 1;
    }
    t = fun(n - 1, fp);
    f = t + *fp;
    *fp = t;
    return f;
}

int main()
{
    int x = 15;
    printf("%d\n", fun(5, &x));

    return 0;
}

import java.io.*;

class GFG {
    static int fp = 15;
    static int fun(int n)
    {
        int t, f;

        if (n <= 2) {
            fp = 1;
            return 1;
        }
        t = fun(n - 1);
        f = t + fp;
        fp = t;
        return f;
    }
    public static void main(String[] args)
    {
        System.out.println(fun(5));
    }
}
// This code is contributed by shubhamsingh10

fp = 15


def fun(n):
    global fp
    if (n <= 2):
        fp = 1
        return 1

    t = fun(n - 1)
    f = t + fp
    fp = t
    return f

# Driver code


print(fun(5))

# This code is contributed by shubhamsingh10

using System;

class GFG {
    static int fp = 15;
    static int fun(int n)
    {
        int t, f;

        if (n <= 2) {
            fp = 1;
            return 1;
        }
        t = fun(n - 1);
        f = t + fp;
        fp = t;
        return f;
    }

    static public void Main() { Console.Write(fun(5)); }
}
// This code is contributed by shubhamsingh10

<script>
//Javascript Implementation
var fp = 15;
function fun( n )
{
    var t, f;
 
    if ( n <= 2 )
    {
        fp = 1;
        return 1;
    }
    t = fun ( n - 1 );
    f = t + fp;
    fp = t;
    return f;
}

// Driver Code

document.write(fun(5));

// This code is contributed by shubhamsingh10

</script>

5

Time complexity: O(n)
Auxiliary Space: O(1) 

The program calculates n-th Fibonacci Number. The statement t = fun ( n-1, fp ) gives the (n-1)th Fibonacci number and *fp is used to store the (n-2)th Fibonacci Number. The initial value of *fp (which is 15 in the above program) doesn't matter. The following recursion tree shows all steps from 1 to 10, for the execution of fun(5, &x). 

                              (1) fun(5, fp)
                              /           \
                         (2) fun(4, fp)   (8) t = 3, f = 5, *fp = 3
                         /          \
                   (3) fun(3, fp)    (7) t = 2, f = 3, *fp = 2
                  /            \
              (4) fun(2, fp)   (6) t = 1, f = 2, *fp = 1
             /   
      (5) *fp = 1   

Question 2: Predict the output of the following program. 

#include <iostream>
using namespace std;
void fun(int n)
{
    if (n > 0) {
        fun(n - 1);
        cout << n << " ";
        fun(n - 1);
    }
}

int main()
{
    fun(4);
    return 0;
}

// This code is contributed by shubhamsingh10

#include <stdio.h>

void fun(int n)
{
    if (n > 0) {
        fun(n - 1);
        printf("%d ", n);
        fun(n - 1);
    }
}

int main()
{
    fun(4);
    return 0;
}

import java.util.*;

class GFG {
    static void fun(int n)
    {
        if (n > 0) {
            fun(n - 1);
            System.out.print(n + " ");
            fun(n - 1);
        }
    }

    public static void main(String[] args) { fun(4); }
}
// This code is contributed by Shubhamsingh10

def fun(n):

    if(n > 0):
        fun(n - 1)
        print(n, end=" ")
        fun(n - 1)

# driver code


fun(4)

# This code is contributed by shubhamsingh10

using System;

class GFG {
    static void fun(int n)
    {
        if (n > 0) {
            fun(n - 1);
            Console.Write(n + " ");
            fun(n - 1);
        }
    }

    static public void Main() { fun(4); }
}

// This code is contributed by shubhamsingh10

<script>
function fun(n)
{
    if(n > 0)
        fun(n - 1);
        document.write(n+" ")
        fun(n - 1); 
}

// driver code
fun(4)

// This code is contributed by bobby.
</script>

1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 

Time complexity: O(n)
Auxiliary Space: O(1) 
 

                     fun(4)
                   /
                fun(3), print(4), fun(3) [fun(3) prints 1 2 1 3 1 2 1]
               /
           fun(2), print(3), fun(2) [fun(2) prints 1 2 1]
           /
       fun(1), print(2), fun(1) [fun(1) prints 1]
       /
    fun(0), print(1), fun(0) [fun(0) does nothing]

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